document.write( "Question 63423: 1. prove that:
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Algebra.Com's Answer #44141 by Edwin McCravy(20060)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "1. prove that:\r\n" );
document.write( "n²-1000n-> infinity as n-> infinity\r\n" );
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document.write( "We must show that given any M > 0,\r\n" );
document.write( "there exists N > 0,\r\n" );
document.write( "in terms of M, such that\r\n" );
document.write( "when n > N, n²-1000n > M\r\n" );
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document.write( "To do this we work backwards first.  \r\n" );
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document.write( "We must find N in terms of M,\r\n" );
document.write( "such that whenever n > N\r\n" );
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document.write( "n² - 1000n > M  \r\n" );
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document.write( "will always be true.\r\n" );
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document.write( "which means \r\n" );
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document.write( "we must find a value of N, in terms of M,\r\n" );
document.write( "such that whenever n > N\r\n" );
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document.write( "    n(n - 1000) > M\r\n" );
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document.write( "is always true whenever n > N\r\n" );
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document.write( "Now we can see that choosing\r\n" );
document.write( "n > M+1000 will make this \r\n" );
document.write( "always true, since if\r\n" );
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document.write( "n > M+1000  \r\n" );
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document.write( "n(n - 1000) > (M+1000)(M+1000-1000) = M(M+1000) > M\r\n" );
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document.write( "So we take N = M+1000 and we have the preceding inequality. \r\n" );
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document.write( "2. use standard results to show that:\r\n" );
document.write( "(n²-3n+1)/(n+5) ->infinity as n->infinity\r\n" );
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document.write( "We must show that given any M > 0, there exists \r\n" );
document.write( "N > 0, in terms of M, such that\r\n" );
document.write( "when n > N, (n²-3n+1)/(n+5) > M\r\n" );
document.write( "\r\n" );
document.write( "To do this we work backwards.  \r\n" );
document.write( "\r\n" );
document.write( "We must find N in terms of M,\r\n" );
document.write( "such that whenever n > N\r\n" );
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document.write( "(n²-3n+1)/(n+5) > M\r\n" );
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document.write( "This will be true whenever:\r\n" );
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document.write( "(n²-3n+1)/(n+5) - M > 0\r\n" );
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document.write( "and since n>0, this will be true\r\n" );
document.write( "whenever\r\n" );
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document.write( "(n²-3n+1) - M(n+5) > 0\r\n" );
document.write( "n² - 3n + 1 - Mn - 5M > 0\r\n" );
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document.write( "which will be true whenever\r\n" );
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document.write( "n² - (3+M)n + (1-5M) > 0\r\n" );
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document.write( "is true. \r\n" );
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document.write( "Using the quadratic formula,\r\n" );
document.write( "The left side has two zeros\r\n" );
document.write( "        ________ \r\n" );
document.write( "(3+M ± Ö1+26M+M²)/2\r\n" );
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document.write( "and the inequality will\r\n" );
document.write( "always be true when n is\r\n" );
document.write( "greater than the larger\r\n" );
document.write( "zero, which is\r\n" );
document.write( "        ________ \r\n" );
document.write( "(3+M + Ö1+26M+M²)/2\r\n" );
document.write( "  \r\n" );
document.write( "So we take N as the larger\r\n" );
document.write( "zero\r\n" );
document.write( "             ________\r\n" );
document.write( " N = (3+M + Ö1+26M+M²)/2\r\n" );
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document.write( "then we can be sure that\r\n" );
document.write( "when n > N, \r\n" );
document.write( "               \r\n" );
document.write( "n > N, (n²-3n+1)/(n+5) > M\r\n" );
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document.write( "Edwin

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